Question

How do you use the chain rule to differentiate `y=1/(x^4-1)`?

Answer 1

`dy/dx= (-4x^3)/(x^4-1)^2`

Explanation:

We have, `y = 1/(x^4-1)`

So, let `u=x^4-1` then:
`(du)/dx = 4x^3`

And then, `y=1/u = u^-1 => dy/(du)=(-1)u^-2`
`:. dy/(du)=-1/u^2`

The chain rule tells us that; `dy/dx = dy/(du) (du)/dx`

Applying this gives us:
`dy/dx = -1/u^2 4x^3`

`= -1/(x^4-1)^2 4x^3`
`= (-4x^3)/(x^4-1)^2`